Hydrogen peroxide and water contain 5.93% and | vedclass

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(A) In $H_2O_2$,hydrogen is $5.93\%$,so oxygen is $100 - 5.93 = 94.07 \ g$.In $H_2O$,hydrogen is $11.2\%$,so oxygen is $100 - 11.2 = 88.8 \ g$.For a f

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Law of Multiple Proportions

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(A) In $H_2O_2$,hydrogen is $5.93\%$,so oxygen is $100 - 5.93 = 94.07 \ g$.In $H_2O$,hydrogen is $11.2\%$,so oxygen is $100 - 11.2 = 88.8 \ g$.For a fixed mass of oxygen (say $1 \ g$):In $H_2O_2$,$1 \ g$ of oxygen combines with $\frac{5.93}{94.07} = 0.063 \ g$ of hydrogen.In $H_2O$,$1 \ g$ of oxygen combines with $\frac{11.2}{88.8} = 0.126 \ g$ of hydrogen.The ratio of the masses of hydrogen that combine with a fixed mass of oxygen is $0.063 : 0.126$,which simplifies to $1 : 2$.Since this is a simple whole-number ratio,the data illustrates the Law of Multiple Proportions.

Hydrogen peroxide and water contain $5.93\%$ and $11.2\%$ of hydrogen,respectively. Which law is illustrated by this data?

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